# Force - conservative or not?

1. Mar 27, 2015

### Calpalned

1. The problem statement, all variables and given/known data
A particular spring obeys the force law $\vec F= (-kx + ax^3 +bx^4)\vec i$. Is this force conservative? Explain why or why not.

2. Relevant equations
Conservative forces depend on the beginning and endpoints, but not the path taken.

3. The attempt at a solution
The given equation for force clearly depends on $x$, which is distance. Therefore, this force is non-conservative. Additionally, if $\vec F$ depends on $x$, then the graph of F vs dl will vary depending on what value of $x$ is plugged in. Work (which is change in energy) will not be constant, so once again force is not conserved. My answer is incorrect, according to my solutions guide. Where did I mess up?

2. Mar 27, 2015

### ShayanJ

Here's where you messed up:
This is wrong. Conservative forces are forces that their work on a particle in moving it from one point to another, doesn't depend on the path taken and only depends on the two points. This is equivalent to saying that a force is conservative if and only if $\oint \vec F \cdot \vec{dl}=0$ for all closed paths. It can be proved that this is equivalent to $\vec \nabla \times \vec F=0$. So the latter equation is a good way of finding out whether a force field is conservative or not.

3. Mar 27, 2015

### Calpalned

What does the symbol integral with circle mean?

The second part of the question asks me to find the potential energy function, which is the integral of $\vec F$. If force is conservative, then the potential energy function is equal to zero right?

4. Mar 27, 2015

### ShayanJ

It means the line integral is taken over a closed path.
No. To obtain the potential energy from the force field, you should use $\phi(\vec r)=-\int_{\vec{r}_{ref}}^{\vec r} \vec F \cdot \vec{dl}$ where $\phi(\vec r _{ref})$ is defined to be zero, i.e. the point where potential becomes zero is arbitrary.

5. Mar 27, 2015

### mattt

In one dimension (as it is this case) a force $\vec{F}$ is conservative if and only if there is a function $\phi(x)$ such that $\vec{F} = -\frac{d}{dx}\phi(x) \vec{i}$

In your example the force is a continuous function of the position $x$ so the above is satisfied (i.e. the force is conservative).